EYE-REFRACTIVE ERRORS

Transcription

1 VISUAL OPTICS LABORATORY EYE-REFRACTIVE ERRORS Prof.Dr.A.Necmeddin YAZICI GAZİANTEP UNIVERSITY OPTİCAL and ACOUSTICAL ENGINEERING DEPARTMENT 1

2 2 REDUCED EYE The refractive state of a given eye is determined by the values of the individual refractive components of the eye and their relationships with one another. These refractive components are corneal refracting power, anterior chamber depth, lens refracting power, and the axial length of the eye. The values of these refractive components are determined both by their radii of curvature (in the case of the cornea and the lens) and by their indices of refraction. For the healthy eye, the indices are thought to show little variation from one eye to another; but the radii of curvature of the cornea and the lens, as well as the anterior chamber depth and the axial length of the eye, are known to vary widely from one eye to another. It is convenient to characterize the refractive components of the eye, together with its focal points, principal points, and nodal points, in terms of what is known as a schematic eye. Such "paper and pencil" eyes are useful in performing a number of calculations relative to the optics of the eye, and are particularly useful for calculations involving magnification.

3 REDUCED EYE There are three major refracting interfaces to be considered in the eye, the anterior (önde) corneal surface and the two surfaces of the lens. The effect of the posterior corneal surface is very small compared with these three as the difference in refractive index between corneal stroma and aqueous is not large. In order to calculate the cardinal points (dört nokta), the radii of curvature, and distances separating the refracting surfaces must also be known. There is some physiological variation in anatomical measurement and these measurements give the optical constants of the eye. Note that the nodal points, via which rays of light pass undeviated, are removed from the principal points, which lie at the intersection of the principal planes with the principal axis (Figure). Figure: The schematic eye. 3

4 Reduced Eye 4 The physical and optical values of a typical human eye have been summarized in below table. The optics of the eye can be reduced to a rather simplified representation referred to as the reduced eye. Below figure shows a diagram of this reduced eye; which is a single refracting surface, bounded by an index of refraction n of 4/3 ( 1.333), with a center of curvature located at point N. The center of curvature of the refracting surface is referred to as the nodal point.

5 5 Reduced Eye Point V serves as the vertex of the surface. The line passing through the vertices V and V' serves as both the visual axis (or the line of sight) and the optical axis of the reduced eye. The axial length of the reduced eye (distance from the surface to the retina) is mm and the refractive power is D, both are close approximations to the real eye. An index of 4/3 (or 1.333) the index of water was chosen because it approximates the index of the fluids of the eye. A goal is to have the reduced eye properly focused. Therefore, the secondary focal distance also needs to be mm so that the image of a distant object can be focused on the retina. The focal length f' of the eye is measured from the vertex V to the secondary focal point F'.

6 Reduced Eye The radius of the refracting surface of the reduced eye is calculated as follows r = m = 5.56 mm The surface s refractive power or reduced eye can be calculated as follows: The image distance is calculated as As a result, a reduced eye consists of a single spherical refracting surface with a radius of curvature of 5.55 mm that separates air from aqueous, which is assumed to have an index of refraction of The eye has a refractive power of D and an axial length of mm. There is a single nodal point located at the center of curvature of the refracting surface. 6

7 Reduced Eye The principal planes are coincident with the refracting surface, and the nodal point is located at its center of curvature. The primary and secondary focal lengths are and mm, respectively. Figure : The reduced eye consists of a single refracting surface (with a radius of curvature of mm) that separates air and aqueous. Figure : An infinitely distant object is imaged on the retina of the reduced eye. 7

8 Reduced Eye Construction of Retinal Image In reduced eye, the retinal image size is directly related to the angle subtended by an object at the nodal point. The retinal image size h = tan(α) f 1 When an object of given size approaches the eye, it subtends a greater visual angle and thus appears larger. If an object subtends an angle θ at the eye, the height h of the image is given by P (principal plane), R (retina), N (nodal point), F a (anterior focus), F 2 (second principal focus) Figure: Reduced eye image formation. 8

9 Reduced Eye Construction of Retinal Image 9 EXAMPLE: Find the size of the image of a mountain 3 miles high, in the eye of an observer located 12 miles from the mountain. SOLUTION From below figure, we see that the mountain subtends an angle θ where tan θ = 3 mi/12 mi = 1/4. Applying the following equation, we obtain the size of the image on the observer s retina. This example shows how even a very large object, filling a large part of the field of view, produces a very small image on the retina.

10 10 Reduced Eye Construction of Retinal Image EXAMPLE: Show that light from a distant object point on the eye s optical axis forms an image on the retina when the eye is relaxed and has the dimensions and curvatures given in below figure. In a reduced eye, we can assumed that there are only three surfaces to be considered: (1) air to aqueous humor; (2) aqueous humor to lens; (3) lens to vitreous humor.

11 Reduced Eye Construction of Retinal Image 11 EXAMPLE (Cont.): (1) We apply the below equation, using an infinite object distance at the first surface, a convex surface from air (n = 1.00) to aqueous humor (n = 1.34), with a radius of curvature R = 7.8 mm. The first image point P 1, 31 mm behind the cornea. (2) The point P1 serves as a virtual object 27 mm behind the second surface a convex surface from aqueous humor (n = 1.34) to lens (n = 1.42), with R = 10 mm. We treat the object distance as negative and again apply the following equation The second image point P 2 is 25 mm behind the front surface of the lens.

12 Reduced Eye Construction of Retinal Image 12 EXAMPLE (Cont.): (3) Point P 2 serves as a virtual object 21 mm behind the third surface a concave surface from lens (n = 1.42) to vitreous humor (n = 1.34), with R = 6.0 mm. Applying below equation once again, we find This is exactly the distance from the lens to the retina, showing that the final image is located on the retina.

13 Reduced Eye Construction of Retinal Image 13 Example: Size of Image on Retina What is the size of the image on the retina of a cm diameter human hair, held at arm s length (60.0 cm) away? Take the lens-to-retina distance to be 2.00 cm. We want to find the height of the image h i, given the height of the object is h o = cm. We also know that the object is 60.0 cm away, so that d o = 60.0 cm. For clear vision, the image distance must equal the lens-to-retina distance, and so d i = 2.00 cm.

14 Visual Acuity 14 Visual acuity is the eye s ability to see fine detail. A measure of acuity is the minimum angle subtended by two points that are barely able to be resolved by the eye, that is, seen as separate points (see figure). This minimum angle, which depends on the individual, is easy to measure. That angle, θ min, measured in radians, is the ratio of the distance between the dots (1 mm) to the distance d of the dots from the eye Under conditions of good illumination with white light, the distance d might be 2 m, for example.

15 Visual Acuity 15 EXAMPLE: What is the minimum distance between two points that you are able to resolve from a distance of 100 m, if you have normal visual acuity? Could you recognize a familiar face 100 m away? SOLUTION The angle θ min subtended by these points is 5x10 4 radians. Using below figure, we can express the separation s of the points in terms of θ min. At a distance of 100 m you will not be able to distinguish details smaller than 5 cm. Thus you may not be able to identify a familiar face at this distance

16 REFRACTIVE ERRORS 16 Refractive errors occur when abnormalities of the eye prevent the proper focus of light on the retina. Emmetropia (göz merceğinin normal oluşu) refers to an eye free of refractive errors. If the secondary focal point fails to coincide with the retina, a refractive error is produced, creating a condition called ametropia. The state of ametropia can be classified as either axial in nature if it occurs because the axial length differs from that of the reduced eye (i.e., the eye is too long or short). Or as refractive in nature if it occurs because the secondary focal length differs from that of the reduced eye (i.e., the power is too strong or weak).

17 17 REFRACTIVE ERRORS A normal eye forms an image on the retina of an object at infinity when the eye is relaxed. Two common types of refractive errors are myopia (or nearsightedness) and hyperopia (or farsightedness). In the myopic (nearsighted) eye, the eyeball is too long from front to back in comparison to the radius of curvature of the cornea (or the cornea is too sharply curved), and rays from an object at infinity are focused in front of the retina. The most distant object for which an image can be formed on the retina is then nearer than infinity.

18 18 REFRACTIVE ERRORS The hyperopic (farsighted) eye has the opposite problem: The eyeball is too short or the cornea is not curved enough, and the image of an infinitely distant object is behind the retina. The myopic eye produces too much convergence in a parallel bundle of rays for an image to be formed on the retina; the hyperopic eye produces not enough convergence.

19 Refractive Errors 19 Myopia and hyperopia are spherical refractive errors that can be corrected with spherical lenses. By definition, a spherical lens has the same refractive power in all of its meridians. As can be seen in below figure, whether it is measured across the horizontal meridian, the vertical meridian, or anywhere between these two meridians, the dioptric power is the same. Figure : A spherical lens has the same power in all of its meridians.

20 Refractive Errors 20 Another common form of ametropia is astigmatism. Astigmatism refers to a defect in which the surface of the cornea is not spherical, but is more sharply curved in one plane than another. As a result, horizontal lines may be imaged in a different plane from vertical lines. Astigmatism may make it impossible, for example, to focus clearly on the horizontal and vertical bars of a window at the same time. FIGURE: (a) An uncorrected astigmatic eye.

21 Refractive Errors Astigmatism is not a spherical refractive error and cannot be fully corrected with a spherical lens, but it can be corrected with what is referred to as a cylindrical lens. For example, suppose the curvature of the cornea in a horizontal plane is correct for focusing rays from infinity on the retina, but the curvature in the vertical plane is not great enough to form a sharp retinal image. Then, when a cylindrical lens with its axis horizontal is placed before the eye, the rays in a horizontal plane are unaffected, but the additional divergence of the rays in a vertical plane causes these to be sharply imaged on the retina, as shown in Fig. FIGURE: (b) Correction of the astigmatism by a cylindrical lens. 21

22 Refractive Errors 22 A cylindrical lens has maximum dioptric power in one meridian, while the orthogonal (perpendicular) meridian has no dioptric power. Figure shows a glass cylindrical lens. This cylinder is positioned so that its axis is horizontal. Along the axis, the cylinder is flat the radius of curvature is infinity and there is no dioptric power. Figure : This glass cylinder has no power along the horizontal meridian and D of power along the vertical meridian. Perpendicular to the axis is the power meridian in which the cylinder has its maximum power ( D). When the correction involves both astigmatism and myopia or hyperopia, there are three numbers for each lens: one for the spherical power, one for the cylindrical power, and an angle to describe the orientation of the cylinder axis.

23 Refractive Errors All these defects can be corrected by the use of corrective lenses (eyeglasses or contact lenses) or, in recent years, by refractive surgery in which the cornea itself is reshaped. The near point of either a presbyopic or a hyperopic eye is farther from the eye than normal. To see an object clearly at normal reading distance (often assumed to be 25 cm), such an eye needs an eyeglass lens that forms a virtual image of the object at or beyond the near point. This can be accomplished by a converging (positive) lens, as shown in Fig. FIGURE: (a) An uncorrected farsighted (hyperopic) eye. (b) A positive lens gives the extra convergence needed to focus the image on the retina. 23

24 24 Refractive Errors In effect, the lens moves the object farther away from the eye, to a point where a sharp retinal image can be formed. Similarly, eyeglasses for myopic eyes use diverging (negative) lenses to move the image closer to the eye than the actual object, as shown in Fig. FIGURE: (a) An uncorrected nearsighted (myopic) eye. (b) A negative lens spreads the rays farther apart to compensate for the eye s excessive convergence.

25 Refractive Errors 25 The far point (FP) of an eye is the position of an object such that its image falls on the retina of the relaxed eye, i.e. in the absence of accommodation. Figure: The far point in emmetropia is infinity. The distance of the far point from the principal plane of the eye is denoted by r, which according to sign convention carries a negative sign in front of the principal plane and a positive sign behind the principal plane. Figure: The far point in hypermetropia is virtual, as only converging light can be focused on the retina. Figure: The far point in myopia lies a finite distance in front of the eye.

26 Refractive Errors 26 In practice, the correcting lens in ametropia (myopia and hyperopia) is usually held in spectacles. The lens is, therefore, some distance in front of the principal plane of the eye. The power of the lens necessary to correct a specific degree of ametropia must therefore be adjusted so that the far point and the focus of the lens still coincide. If a correcting lens is moved either towards or away from the eye, its vergence power at the principal plane of the eye changes. The focus of the lens and the far point of the eye no longer coincide (see below figures). It can also be seen from the diagrams that, on moving either a convex or a concave lens away from the eye, the image is moved forward. The purpose of the correcting lens in ametropia is to deviate parallel incident light so that it appears to come from the far point in myopia or to be converging towards the virtual far point in hypermetropia. The light will then be brought to a focus by the eye on the retina. Thus the far point of the eye must coincide with the focal point of the lens.

27 Refractive Errors Figure: Diagram showing the change in effectivity of a convex lens on moving it away from the eye. 27

28 Refractive Errors FigUre: Diagram showing the change in effectivity of a concave lens on moving it away from the eye. 28

29 Refractive Errors 29 The focal length, f, of the correcting lens is approximately equal to the distance, r, of the far point from the principal plane when the correcting lens is close to the principal plane. Thus the power of lens, F, required is where F is the power of the lens in dioptres; f is the focal length of the lens in metres; and r is the distance of the far point from the principal plane in metres. The reciprocal of the far point distance r, in metres, is symbolised by R, expressed in dioptres. R = 1/r R is known as the static refraction or the ametropic error.

30 Refractive Errors 30 A general formula applies to find power of lenses of both convex and concave lenses. Suppose a lens of focal length f 1 at a given position in front of the ametropic eye corrects the refractive error. Then a different lens of focal length (f 1 d) is required when the correction is moved a distance d towards or away from the eye. The value of d is positive if the lens is moved towards the eye, and negative if moved away from the eye. The usual sign convention applies to the lens. where F 2 is the power of lens in dioptres required at the new position; f 1 is the focal length in metres of the original lens; and d is the distance moved in metres. Mathematically the above formula can also be expressed as where F 1 is the dioptric power of the original lens.

31 Refractive Errors 31 Example 1 Refraction shows that an aphakic patient requires a D lens at back vertex distance (BVD) which is the distance between the back of the lens and the cornea is 15 mm. He needs a contact lens (F 2 ) F 2 = D Example 2 Likewise a high myope whose spectacle correction is 10.0 D at BVD 14 mm requires a contact lens (F 2 ) F 2 =-8.75 D

32 Myopia Myopia, also known as near-sightedness, occurs if the eye is longer than normal or the curve of the cornea is too steep, causing light rays focus in front of the retina (the second principal focus lies in front of the retina). Either the optical elements of the eye are too strong, or the axial length of the eye is too long. It is called as axial myopia. Alternatively, the eye may be of normal length, but the dioptric power may be increased. This is called refractive or index myopia. Patients with myopia are able to see objects at near, but distant objects appear blurred. FIGURE: Myopic focus (in front of the retina). 32

33 Myopia The distinction between axial and refractive ametropia may best be understood by considering the eye as having two "lengths": the axial length (the physical length of the eyeball) and the focal length of the eye's optical system, that is, the distance from the secondary principal plane to the secondary focal plane of the eye. These two "lengths" are shown in below figure. If the axial length of the eye is shorter or longer than that of the schematic eye, while the secondary focal length corresponds to that of the schematic eye, the eye will have axial hyperopia or myopia, respectively. If the axial length of the eye in question corresponds to that of the schematic eye but the focal length of the optical system is either longer or shorter than that of the schematic eye, the eye will have refractive hyperopia or myopia, respectively. FIGURE: The axial length (physical length) of the eyeball as compared to the secondary focal length of the eye's optical system. 33

34 Myopia 34 Clear vision can be restored to most myopes through the use of minus-powered lenses. In the myopic eye the image falls in front of the retina. The purpose of the correcting concave lens is to take the image back on to the retina. When the correcting lens is moved further away from the eye, the image moves forward again. Thus the effectivity of the lens is said to be reduced. Therefore, in this position, a stronger concave lens is needed to throw the image on to the retina. FIGURE: Minus lens correction. A) Parallel rays of light from a distant object focus in front of the retina of the myopic eye. B) Rays of light diverging from the far point M R of the eye focus on the retina. C) The corrective minus lens diverges parallel rays from an object at infinity to form a virtual image at F', which corresponds to the far point M R of the eye. This image now becomes a real object for the eye that is conjugate with the retina, and forms a clear image upon it.

35 Myopia Figure: Correction of myopia. 35

36 Myopia 36 EXAMPLE: Correcting for nearsightedness The far point of a certain myopic eye is 50 cm in front of the eye. What lens should be used to focus sharply an object at infinity? (Assume that the distance from the lens to the eye is negligible.) The far point of a myopic eye is nearer than infinity. To see clearly objects that are beyond the far point, such an eye needs a lens that forms a virtual image of the object no farther from the eye than the far point. We assume that the virtual image is formed at the far point. Then, when s =, we want s to be -50 cm.

37 Myopia 37 EXAMPLE (Cont.): Correcting for nearsightedness From the basic thin-lens equation, Because all rays originating at an object distance of infinity are parallel to the axis of the lens, the image distance equals the focal length f. We need a diverging lens with focal length f = - 50 cm = m. The power is - 1/ (0.50 m) = diopters.

38 Myopia EXAMPLE: A Corrective Lens for Nearsightedness A particular nearsighted patient can t see objects clearly when they are beyond 25 cm (the far point of the eye). (a) What focal length should the prescribed contact lens have to correct this problem? (b) Find the power of the lens, in diopters. Neglect the distance between the eye and the corrective lens. The purpose of the lens in this instance is to take objects at infinity and create an image of them at the patient s far point. (a) Find the focal length of the corrective lens, apply the thin-lens equation for an object at infinity and image at 25.0 cm: (b) Find the power of the lens in diopters: The focal length is negative, consistent with a diverging lens. Notice that the power is also negative and has the same numeric value as the sum on the left side of the thin-lens equation. 38

39 Myopia 39 EXAMPLE: A Case of Nearsightedness a-) A particular nearsighted person is unable to see objects clearly when they are beyond 2.5 m away (the far point of this particular eye). What should the focal length be in a lens prescribed to correct this problem? Solution The purpose of the lens in this instance is to move an object from infinity to a distance where it can be seen clearly. This is accomplished by having the lens produce an image at the far point. From the thin-lens equation, we have As you should have suspected, the lens must be a diverging lens (one with a negative focal length) to correct nearsightedness. b-) What is the power of this lens? P =1/f = 0.40 diopter.

40 Myopia 40 Example: Correcting Nearsightedness What power of spectacle lens is needed to correct the vision of a nearsighted person whose far point is 30.0 cm? Assume the spectacle (corrective) lens is held 1.50 cm away from the eye by eyeglass frames. You want this nearsighted person to be able to see very distant objects clearly. That means the spectacle lens must produce an image 30.0 cm from the eye for an object very far away. An image 30.0 cm from the eye will be 28.5 cm to the left of the spectacle lens. Therefore, we must get d i = 28.5 cm when d o. The image distance is negative, because it is on the same side of the spectacle as the object. Since d i and d o are known, the power of the spectacle lens can be found using The negative power indicates a diverging (or concave) lens, as expected. The spectacle produces a case 3 image closer to the eye, where the person can see it. If you examine eyeglasses for nearsighted people, you will find the lenses are thinnest in the center. Additionally, if you examine a prescription for eyeglasses for nearsighted people, you will find that the prescribed power is negative and given in units of diopters.

41 Myopia 41 Example: If a myopic eye that is corrected with a 6.68 D lens placed in the plane of the cornea. what power spectacle lens is required to correct the refractive error with a spectacle lens that is located mm in front of the eye? Figure : This refractive error can be corrected with (A) a 6.68 D contact lens. To correct the refractive error, the secondary focal point of the correcting lens must be coincident with far point. The far point is cm anterior to the cornea.

42 Myopia Example (Cont.): If the correcting lens is mm anterior to the cornea, its secondary focal length must be cm [( cm) ( 1.50 cm) = cm]. The power of a minus lens with a secondary focal length of cm is This myopic eye can be corrected with either a 6.68 D contact lens or a 7.42 D spectacle lens at a vertex distance of mm. Both have the same effective power. Figure: This refractive error can be corrected with a 7.42 D spectacle lens. 42

43 43 Myopia EXAMPLE: (a) Find the optical power of a lens necessary to correct an eye with a far point of 50 cm. Neglect the distance from the lens to the cornea. (b) If the eye s near point is 10 cm, what is the eye s corrected near point? SOLUTION (a) A diverging or negative lens with a focal length of magnitude 50 cm is required. f = -50 cm = m Thus the power of the lens is (b) An object will be at the corrected near point when the image produced by the lens is at the eye s actual near point. Here the eye s near point is 10 cm in front of the lens. Thus we must find the object distance s for which the lens produces a virtual image 10 cm in front of the lens, that is, for which s = 10 cm. Applying the thin lens equation, we find The original range of vision for this myopic eye was 10 cm to 50 cm. After correction, the range becomes 13 cm to.

44 Myopia 44 Example: Correcting myopia Martina has myopia. The far point of her left eye is 200 cm. What prescription lens will restore normal vision? Normal vision will allow Martina to focus on a very distant object. In measuring distances, we ll ignore the small space between the lens and her eye. Because Martina can see objects at 200 cm with a fully relaxed eye, we want a lens that will create a virtual image at position s = -200 cm (negative because it s a virtual image) of a distant object at s = cm. From the thin-lens equation, Thus the prescription is for a lens with power P = -0.5 D. Myopia is always corrected with a diverging lens.

45 Hyperopia Hyperopia, also known as far-sightedness, occurs if the eye is too short or the curve of the cornea is too flat, causing light rays to focus behind the retina. Patients with hyperopia are able to see objects at distance, but near objects appear blurred. Either the optical elements of the eye are too weak (refractive hypermetropia), or the axial length of the eye is too short (axial hypermetropia). Mildly (hafifçe) hyperopic patients may be able to see clearly at near without correction by using accommodation to compensate. Clear vision can be restored to most hyperopes through the use of plus-powered lenses. FIGURE Hyperopic focus (behind the retina). 45

46 Hyperopia 46 Hypermetropia is also classified into manifest hypermetropia and latent hypermetropia In the uncorrected hypermetropic eye the image falls behind the retina. The purpose of the correcting convex lens is to bring the image forward on to the retina. When the correcting lens is moved further away from the eye the image is brought still further forward. Thus the effectivity of the lens is said to be increased. Therefore in this position a weaker convex lens throws the image onto the retina and corrects the hypermetropia. FIGURE: Plus lens corrections. A) Parallel rays of light from a distant object focus behind the retina of the hyperopic eye. B) Rays of light converging to the far point M R of the eye focus on the retina. C) The corrective plus lens converges parallel rays from an object at infinity to form a real image at F', which corresponds to the far point M R of the eye. This image now becomes a virtual object for the eye that is conjugate with the retina, and forms a clear image upon it.

47 Hyperopia Figure: Correction of hypermetropia. 47

48 Hyperopia EXAMPLE: Correcting for farsightedness The near point of a certain hyperopic eye is 100 cm in front of the eye. What lens should be used to enable the eye to see clearly an object that is 25 cm in front of the eye? (Neglect the small distance from the lens to the eye.) We want the lens to form a virtual image of the object at a location corresponding to the near point of the eye, 100 cm from the lens. That is, when s = 25 cm, we want s to be cm. From the basic thin-lens equation We need a converging lens with focal length f = 33 cm. The corresponding power is 1/ (0.33 m) or +3.0 diopters. 48

49 Hyperopia 49 Example 26.4 Correcting Farsightedness What power of spectacle lens is needed to allow a farsighted person, whose near point is 1.00 m, to see an object clearly that is 25.0 cm away?assume the spectacle (corrective) lens is held 1.50 cm away from the eye by eyeglass frames. When an object is held 25.0 cm from the person s eyes, the spectacle lens must produce an image 1.00 m away (the near point). An image 1.00 m from the eye will be 98.5 cm to the left of the spectacle lens because the spectacle lens is 1.50 cm from the eye. Therefore, d i = 98.5 cm. The image distance is negative, because it is on the same side of the spectacle as the object. The object is 23.5 cm to the left of the spectacle, so that d o = 23.5 cm. The power of the spectacle lens can be found using The positive power indicates a converging (convex) lens, as expected. The convex spectacle produces a case 2 image farther from the eye, where the person can see it. If you examine eyeglasses of farsighted people, you will find the lenses to be thickest in the center. In addition, a prescription of eyeglasses for farsighted people has a prescribed power that is positive.

50 Hyperopia 50 Example: Correcting hyperopia Sanjay has hyperopia. The near point of his left eye is 150 cm. What prescription lens will restore normal vision? Normal vision will allow Sanjay to focus on an object 25 cm away. In measuring distances, we ll ignore the small space between the lens and his eye. Because Sanjay can see objects at 150 cm, using maximum accommodation, we want a lens that creates a virtual image at position s = -150 cm (negative because it s a virtual image) of an object held at s = 25 cm. From the thin-lens equation, 1/f is the lens power, and m -1 are diopters. Thus the prescription is for a lens with power P = 3.3 D. Hyperopia is always corrected with a converging lens.

51 Hyperopia 51 EXAMPLE: Prescribing a Corrective Lens for a Farsighted Patient The near point of a patient s eye is 50.0 cm. (a) What focal length must a corrective lens have to enable the eye to see clearly an object 25.0 cm away? Neglect the eye lens distance. (b) What is the power of this lens? (c) Repeat the problem, taking into account the fact that, for typical eyeglasses, the corrective lens is 2.00 cm in front of the eye. Solution (a) Find the focal length of the corrective lens, neglecting its distance from the eye. Apply the thin-lens equation: Substitute p = 25.0 cm and q = cm (the latter is negative because the image must be virtual) on the same side of the lens as the object: The focal length is positive, corresponding to a converging lens.

52 Hyperopia 52 Example (Cont.): (b) The power is the reciprocal of the focal length in meters: (c) Repeat the problem, noting that the corrective lens is actually 2.00 cm in front of the eye. Substitute the corrected values of p and q into the thin-lens equation: Compute the power: Notice that the calculation in part (c), which doesn t neglect the eye lens distance, results in a difference of 0.26 diopter.

53 53 Example: A hyperopic patient requires a corrective lens of D in the plane of the cornea (i.e., a contact lens whose power is +2.82D). What power spectacle lens is required to correct this refractive error? Assume a vertex distance of mm. Figure : This refractive error can be corrected with a D contact lens The secondary focal length of the spectacle lens must be cm ( cm = cm). The power of this spectacle lens is

54 54 Example (Cont.): This hyperope can be corrected with either a D contact lens or a D spectacle lens. The contact lens has a higher power because it must have a shorter focal length if its secondary focal point to be coincident with the eye s far point. Figure : This refractive error can be corrected with a D spectacle lens. Question: Your refraction at a vertex distance of mm results in a spectacle prescription of 9.00 D in each eye. If the patient wishes to be fit with contact lenses, what power should they be?

55 55 EXAMPLE: Find the power of a lens designed for reading purposes to correct an eye with a near point of 75 cm. Assume a standard reading distance of 25 cm. Neglect the distance from the corrective lens to the cornea. SOLUTION A converging or positive lens is required, with a power such that when an object is placed 25 cm in front of the lens, the lens forms a virtual image as shown in below figure. The image distance s = 75 cm, so that this image serves as an object for the eye at the eye s near point. Applying the thin lens equation, we find Figure: A hyperopic eye corrected for viewing a close object. The image produced by the lens is at the eye s near point.

56 56 EXAMPLE: An uncorrected myope has a far point of accommodation located 26.5 cm in front of the corneal plane. What would be the required power of the correcting lens if the lens is to be located (a) 15 mm in front of the corneal apex; (b) 10 mm in front of the corneal apex? (see below figure). (a) The power of the correcting lens placed 15 mm from the corneal apex would be the reciprocal of the secondary focal length of the lens. (b) The focal length of the lens, if located 10 mm in front of the corneal apex, would be or FIGURE: The effective power of a lens: the correction of a myopic eye with lenses located at two different distances from the corneal apex.

57 EXAMPLE: An uncorrected hyperopic eye has a far point of accommodation located 18.5 cm behind the corneal apex. What would be the required power of a correcting lens located (a) 15 mm in front of the corneal apex; (b) 20 mm in front of the corneal apex? (see figure). (a) The power of the correcting lens, placed 15 mm from the corneal apex, would be the reciprocal of the secondary focal length of the lens. (b) The focal length of the correcting lens, for a 20-mm fitting distance, would be As an effectivity problem, assigning to d a value of -5 mm, FIGURE: The effective power of a lens: the correction of a hyperopic eye (B) with lenses located at two different distances from the corneal apex. 57

58 58 Example: Calculate the power of the eye when viewing objects at the greatest and smallest distances possible with normal vision, assuming a lens-to-retina distance of 2.00 cm (a typical value). For clear vision, the image must be on the retina, and so d i = 2.00 cm here. For distant vision, d o, and for close vision, d o = 25.0 cm, as discussed earlier. For distant vision; For close vision,

59 59 EXAMPLE: A farsighted person has a near point of 200 cm.what kind of glasses does this person need to be able to read a newspaper held at 25 cm from the eyes? SOLUTION: The object distance is 25 cm (we ignore the small difference between distances measured from the eye and from the glasses). Since the nearest point at which the person can see distinctly is 200 cm away, the lens must form an image of the newspaper at this distance that is, s = cm. The lens equation then tells us The inverse of the focal length in meters; thus for the farsighted person of this example, reading glasses of strength 1/0.29 = diopters would permit reading at 25 cm.

60 Astigmatism 60 An even more common type of refractive error is astigmatism. Astigmatism occurs when the cornea has an oblong (boyu eninden fazla), football-like shape in one or more directions (or axes) causing light rays to focus on more than one point on the retina. This refractive error occurs when rays of light do not come to a focal point, but instead form two focal lines after refraction by the optical elements of the eye. The refractive power of the astigmatic eye varies in different meridians. Astigmatism is compensated for using cylinder powered lenses along the appropriate axis.

61 Astigmatism 61 There are five distinct types of astigmatism, depending upon the combination of refractive errors in the principal meridians of the eye: Mixed astigmatism (hyperopic in one meridian, myopic in the other), compound myopic astigmatism (myopic in both meridians), simple myopic astigmatism (myopic in one meridian, emmetropic in the other), compound hyperopic astigmatism (hyperopic in both meridians), and simple hyperopic astigmatism (hyperopic in one meridian, emmetropic in the other).

62 Astigmatism 62 These five types of astigmatism, and their corresponding focal lines, are shown in below figure. FIGURE: Types of astigmatism: mixed astigmatism (one focal line in front of and one behind retina), compound myopic astigmatism (both focal lines in front of retina), simple myopic astigmatism (one focal line in front of and one on retina), compound hyperopic astigmatism (both focal lines behind retina), and simple hyperopic astigmatism (one focal line behind and one on retina). The actual orientation of these focal lines will vary but they will always be at right angles to one another (90 apart).

63 Astigmatism Retina a = Compound hypermetropic astigmatism rays in all meridians come to a focus behind the retina. Retina b = Simple hypermetropic astigmatism rays in one meridian focus on the retina, the other focus lies behind the retina. Retina c = Mixed astigmatism one line focus lies in front of the retina, the other behind the retina. Retina d = Simple myopic astigmatism one line focus lies on the retina, the other focus lies in front of the retina. Retina e = Compound myopic astigmatism rays in all meridians come to a focus in front of the retina. Figure: Astigmatism; classification. Image position relative to retina. 63

64 Astigmatism 64 If the principal meridians are at 90 to each other, this is called regular astigmatism. If the principal meridians are at 90 to each other but do not lie at or near 90 and 180, the term oblique astigmatism is used. If the principal meridians are not at 90 to each other, this is called irregular astigmatism and cannot be corrected by spectacles.

65 Presbyopia 65 As eyes age, the crystalline lens begins to lose elasticity. With the loss of elasticity, the eye loses the ability to accommodate or focus at near. This typically becomes noticeable around 40 years of age. This condition where the crystalline lens is unable to add sufficient power to focus at near is known as presbyopia. The loss of elasticity in the crystalline lens continues until somewhere around the age of 65 when all the elasticity is gone from the lens as is all ability to accommodate. In infancy (bebeklik) the eye is capable of 14 D of accommodation, but by the age of 45 years this has fallen to about 4 D. After the age of 60 years only 1 D or less remains, and part of this is probably due to depth of field, which may be enhanced by senile (bunak-yaşlı) miosis (göz bebeğinin büzülmesi). A person experiencing difficulty and discomfort for near vision due to reduced amplitude of accommodation is said to be presbyopic. Presbyopia can be compensated for through the use of plus-powered lens segments, reading glasses, or magnifying devices.

66 Presbyopia If a patient with 3 D of hypermetropia needs to exert 3D of accommodation to see clearly at infinity, therefore, 7D of accommodation are needed (3D + 4D) to see clearly at 25 cm (see figures). Conversely a patient with 3D of myopia has a far point at 33 cm. Thus to focus at 25 cm only 1D of accommodation is used. Presbyopia is corrected by adding plus lens power to the patient s distance correction. This additional plus lens power is often in the form of a bifocal add. Figure: Amplitude of accommodation necessary to achieve clear vision at 25 cm in different refractive states. Figre: Decline in amplitude of accommodation with increasing age. 66

67 Presbyopia 67 The extremes of the range over which distinct vision is possible are known as the far point and the near point of the eye. The far point of a normal eye is at infinity. The position of the near point depends on the amount by which the ciliary muscle can increase the curvature of the crystalline lens. The range of accommodation gradually diminishes with age as the crystalline lens loses its flexibility. For this reason, the near point gradually recedes as one grows older. This recession of the near point is called presbyopia; it is the reason that people need reading glasses when they get older, even if their vision is good otherwise. Below table shows the approximate position of the near point for an average person at various ages. TABLE: Receding of near point with age For example, an average person 50 years of age cannot focus on an object closer than about 40 cm.

68 Anisometropi 68 Anisometropia is a condition in which the two eyes have unequal refractive power. When the refraction of the two eyes is different, the condition is known as anisometropia. Antimetropia is an extreme case of anisometropia where one eye is myopic and the other is hyperopic. The unequal refractive states can often lead to diplopia (double vision) or asthenopia (eye strain). Anisometropia can adversely affect the development of binocular vision in infants (süt çocuğu) and children, if there is a large difference between the two eyes. The brain will often suppress the vision of the blurrier (bulanık) eye in a condition called amblyopia (Göz dönüklüğü hastalığı), or lazy (tembel) eye.

69 69 Accommodation The function of accommodation is to bring objects at near into focus. During this process, the crystalline lens becomes more bi-convex; effectively increasing the plus power of the eye. As the lens ages, however, it gradually becomes less flexible, and slowly loses its accommodative power (or amplitude of accommodation).

70 70 Accommodation 1. The far point of distinct vision is the position of an object such that its image falls on the retina in the relaxed eye, i.e. in the absence of accommodation. 2. The near point of distinct vision is the nearest point at which an object can be clearly seen when maximum accommodation is used. 3. The range of accommodation is the distance between the far point and the near point. 4. The amplitude of accommodation is the difference in dioptric power between the eye at rest and the fully accommodated eye. 5. The dioptric power of the resting eye is called its static refraction. 6. The dioptric power of the accommodated eye is called its dynamic refraction.

71 71 Accommodation The amplitude of accommodation is given by the formula A = P - R where A is the amplitude of accommodation in dioptres; P is the dioptric value of the near point distance; and R is the dioptric value of the far point distance. FIGURE: An object 25 cm in front of the eye (0.25 m) produces D of divergence. A) The optical system of an emmetropic eye (with no error) is effectively weak by this amount, with accommodation at rest. B) With D of accommodation, however, the image is brought back into focus.

72 72 Accommodation The vergence relationship for accommodation F FP = L + F A where F FP is the far-point vergence as measured at the cornea, L is the stimulus to accommodation (i.e., the vergence that the object produces at the cornea), and F A is the accommodation, measured in the plane of the cornea, required to focus the object on the retina. Example: An object is located cm in front of the cornea of an emmetropic eye. How much must the eye accommodate for the object to be imaged on the retina? The object distance is cm, making the stimulus to accommodation is 3.00 D [e.g., (1.00)(100)/ cm = 3.00 D]. Since the far point of an emmetropic eye is at infinity, the far-point vergence is 0.00 D. Substituting into the vergence relationship for accommodation, we have F FP = L + F A 0.00 D = 3.00 D + F A F A = D The emmetropic eye must accommodate 3.00 D in order for an object at cm to be imaged on the retina.

73 Accommodation FIGURE: Refractive errors and their corresponding far points of accommodation. For the emmetrope, parallel rays of light focus on the retina; the far-point conjugate to the retina is also at infinity. For the myope, parallel rays of light come to a focus in front of the retina; the far point is a real object point in front of the eye, but within optical infinity. For the hyperope, parallel rays of light come to a focus behind the retina; the far point is a virtual object point lying behind the eye. The total amplitude of accommodation is simply the dioptric difference between the far and near points of accommodation. If both the far and near points of accommodation are known, the amplitude of accommodation A can also be quickly determined, since A = M R - M P. 73

74 As shown in below figure, the far point of accommodation for an emmetropic eye is at infinity (Figure-A); the far point of accommodation for an uncorrected myopic eye is a real object point in front of the eye (Figure-B); and the far point of accommodation for an uncorrected hyperopic eye is a virtual object point located behind the eye (Figure-C). Accommodation FIGURE: Far point of accommodation: A, for an emmetropic eye; B, for an uncorrected myopic eye; C, for an uncorrected hyperopic eye. 74

75 75 Accommodation As shown in Figure, for both an emmetropic eye (Figure-A) and an uncorrected myopic eye (Figure-B), the near point of accommodation is a real object point located in front of the eye, while for an uncorrected hyperopic eye (Figure-C) the near point of accommodation may be either a real object point located in front of the eye or a virtual object point located behind the eye (depending upon both the amount of hyperopia and the amplitude of accommodation). FIGURE: Near point of accommodation: A, for an emmetropic eye; B, for an uncorrected myopic eye; C, for an uncorrected hyperopic eye..

76 76 Amplitude of Accommodation Accommodation may be specified in terms of (a) range or (b) amplitude. The range of accommodation is equal to the distance from the far point of accommodation to the near point of accommodation. Amplitude of accommodation is defined as the dioptric difference between the far and near points of accommodation. For an eye in either the uncorrected or the corrected state, the amplitude of accommodation is given by the expression, where M r S is the linear distance from the far point of accommodation to the spectacle plane, and M p S is the linear distance from the near point of accommodation to the spectacle plane. FIGURE: The range of accommodation: A, emmetropic eye; B, myopic eye.

77 77 Amplitude of Accommodation FIGURE: The range of accommodation: C hyperopic eye.

78 Ocular Accommodation 78 The amount of ocular accommodation required of an eye (see below figure) can be determined by the use of the formula where V d is the vergence of light in the principal plane for an object at a distance of 6 m (assumed to be infinity), and V n is the vergence of light in the principal plane for an object at near distance (usually assumed to be at 40 cm). FIGURE: A, vergence at the principal plane (P) for an object at infinity (V d ); B, vergence at the principal plane (P) for a near object (V n ).

79 79 Accommodation EXAMPLE: Given an emmetropic eye (having its far point at infinity) as shown in below figure, having a near point of accommodation located at a distance of 10 cm in front of the spectacle plane. What are (a) the range and (b) the amplitude of accommodation? (a) The range of accommodation extends from infinity to a distance of 10 cm from the spectacle plane; therefore it is equal to infinity. (b) The amplitude of accommodation FIGURE:. Determination of range and amplitude of accommodation for (A) an emmetropic eye, This example shows that for an emmetropic eye, having its far point of accommodation located at infinity, the range of accommodation is equal to infinity minus a finite number, which is, of course, still infinity. The amplitude of accommodation, on the other hand, is equal simply to the reciprocal of the near point of accommodation.

80 80 Accommodation EXAMPLE: Given an uncorrected myopic eye (Figure) having a far point of accommodation located 50 cm in front of the spectacle plane and a near point of accommodation 10 cm in front of the spectacle plane. What are (a) the range and (b) the amplitude of accommodation? (a) The range of accommodation extends from 50 to 10 cm from the spectacle plane, and is therefore equal to 40 cm. (b) The amplitude of accommodation FIGURE: Determination of range and amplitude of accommodation for an uncorrected myopic eye,

81 Accommodation EXAMPLE: Given an uncorrected hyperopic eye (Figure) having a far point of accommodation located 50 cm behind the spectacle plane and a near point of accommodation located 10 cm in front of the spectacle plane. What are (a) the range and (b) the amplitude of accommodation? (a) The range of accommodation extends from 50 cm behind the spectacle plane to 10 cm in front of the spectacle plane, and is therefore equal to infinity. (b) The amplitude of accommodation FIGURE: Determination of range and amplitude of accommodation for uncorrected hyperopic eye. This example shows that, if an uncorrected hyperopic eye has a finite, or real, near point of accommodation, the range of accommodation (like that of the emmetropic eye) will be "infinite." 81

82 Accommodation EXAMPLE: Given an uncorrected hyperopic eye (Figure) having a far point of accommodation located 10 cm behind the spectacle plane and a near point of accommodation located 25 cm behind the spectacle plane. What are (a) the range and (b) the amplitude of accommodation? (a) The range of accommodation extends from 10 cm behind the spectacle plane to 25 cm behind the spectacle plane, and is therefore equal to 15 cm. (b) The amplitude of accommodation FIGURE: Determination of range and amplitude of accommodation for uncorrected hyperopic eye. It should be understood that, since both the far and near points of the eye in this example are located behind the spectacle plane, it is impossible to place a real object at either of these points. 82

83 83 Accommodation The far and near points of accommodation can be converted into their dioptric equivalents by formula for converting back and forth from distances to diopters. This is familiar formula for vergence: M =1/m Here, m is the distance from the spectacle plane to either the far or near point M of the eye in meters. For presbyopic patients that require both near and distance corrections, at least two distinct focal powers need to be provided. The most common method of providing additional plus power (i.e. the add) at near is with the use of a multifocal lens, which is simply a lens with more than one focal power for different distances. Figure illustrates a typical flattop bifocal lens. The major portion provides distance vision, and the seg (which is short for segment ) provides the add power for near. FIGURE: The flat-top bifocal lens.

84 84 Example: An eye has a real far point of accommodation 50 cm (-0.5 m) in front of the spectacle plane, and a near point 20 cm ( m) in front of it. What are the distance refractive error and the amplitude of accommodation? Distance refractive error is D and the amplitude of accommodation is D. FIGURE: Eye with D of myopia and an amplitude of accommodation of D. The far point of accommodation M R is 50 cm in front of the eye, while the near point M P is 20 cm in front of the eye.

85 85 Example: An eye has a virtual far point of accommodation 50 cm (0.5 m) behind the spectacle plane, and a near point 25 cm (-0.25 m) in front it. What are the distance refractive error and the amplitude of accommodation? Distance refractive error is D and the amplitude of accommodation is D. FIGURE: Eye with D of hyperopia, and an amplitude of accommodation of D. The far point of accommodation M R is 50 cm behind the spectacle plane, while the near point M P is 25 cm in front of it.

86 Spectacle Refraction The power of the correcting lens, specified at the spectacle plane, is termed spectacle refraction. For an eye whose spectacle (see below figure), refraction is D, the secondary focal length of a correcting lens located at the primary principal plane of the eye would be 10 cm cm cm, and the refracting power of the correcting lens, or the ocular refraction, would be FIGURE: For a D hyperope, secondary focal length of the correcting lens, (A) measured from the spectacle plane, is 10 cm, B, measured from the primary principle plane, is 8.5 cm. 86

87 87 EXAMPLE As a more probable example, consider a patient whose distance prescription is OD DS OS DS What is the induced anisometropia at 40 cm for a young patient having a high amplitude of accommodation? For the young patient, we must calculate the values V d and V n for each eye. To calculate V d, we first use the formula L = L+ F. For the right eye, the vergence at distance is given by at the spectacle plane. The image formed by the lens is the object for the eye; so the object distance, l(for the first principal plane of the eye), is given by

88 88 EXAMPLE (Cont.) To find the value of V n, we use the relationship at the spectacle plane. The near-object distance, l, for the principal plane of the eye is given by For the left eye, the vergence at distance is given by at the spectacle plane.

89 89 EXAMPLE (Cont.) The object distance, l (for the first principal plane of the eye), is given by To find the value of V n, measured from the spectacle plane. The near-object distance, l, measured from the principal plane of the eye, is given by

90 90 EXAMPLE (Cont.) The induced anisometropia at 40 cm is equal to the accommodation required by the right eye less the accommodation required by the left eye, or,

91 Spectacle Magnification 91 By comparing the sizes of the object and image (using a ruler), we can also determine the lateral magnification (M L ) produced by the surface. Real images are always inverted and an inverted image is designated by a minus sign preceding the magnification. When an image is erect (virtual image), its magnification is designated by a plus sign. The optical correction of ametropia is associated with a change in the retinal image size. The ratio between the corrected and uncorrected image size is known as the spectacle magnification.

92 Spectacle Magnification 92 Clinically, it is more useful to compare the corrected ametropic image size with the emmetropic image size. This ratio is known as the relative spectacle magnification (RSM). In axial ametropia, if the correcting lens is placed at the anterior focal point of the eye, the image size is the same as in emmetropia.

93 Spectacle Magnification 93 The RSM is unity. However, in axial myopia, if the correcting lens is worn nearer to the eye than the anterior focal point, the image size is increased. The relative spectacle magnification is therefore greater than unity. Contact lenses in axial myopia thus have a magnifying effect. Figure: Relative spectacle magnification. Axial ametropia with correcting lens at anterior focal point of the eye. Ametropic state (solid lines) compared with emmetropia (dotted lines).

94 Spectacle Magnification Figure: Relative spectacle magnification. Axial myopia with correcting lens nearer the eye than the anterior focal point (contact lens). Ametropic state (solid lines) compared with emmetropia (dotted lines). lem is the image size when correction is at the anterior focal point, which equals the emmetropic image size; and lcl is the image size when correction is closer to the eye than the anterior focal point. Convex cylindrical lenses are also employed as reading aids (see figure). The bar-shaped lens which has no refractive power or only a low converging power in its long axis and high converging power in cross-section is laid on a line of print and produces vertical magnification of the letters. Figure: Convex cylindrical magnifying lens. 94

95 Spectacle Magnification In contrast to axial ametropia, the image size in refractive ametropia differs from the emmetropic image size even when the correcting lens is at the anterior focal point of the eye. The image size in refractive hypermetropia is increased, thus the relative spectacle magnification is greater than unity. In refractive myopia the image size is diminished, and thus the relative spectacle magnification is less than unity. Furthermore, in refractive ametropia, if the correcting lens is worn nearer to the eye than the anterior focal point, the image size approaches the emmetropic image size. The relative spectacle magnification thus approaches unity Figure: Relative spectacle magnification. Refractive ametropia with correcting lens at anterior focal point of the eye. Ametropic state (solid lines) compared with emmetropia (dotted lines). 95

96 Example: In the given figure: Emmetropic anterior focal length, F em D = mm, Aphakic anterior focal length F aph D = mm Calculate relative spectacle magnification (RSM). Spectacle Magnification Since rays F em E, and F aph G are parallel, Figure: Correction of aphakia with spectacles at anterior focal point. Ametropic state (solid lines) compared with emmetropia (dotted lines). and Therefore; RSM=1.36 It has been shown earlier that the relative spectacle magnification produced by aphakic spectacle correction is approximately This means that the image produced in the corrected aphakic eye is one third larger than the image formed in an emmetropic eye. This magnification causes the patient to misjudge distances. Objects appear to be closer to the eye than they really are because of the increased visual angle subtended at the eye. 96

97 Spectacle Magnification The field of vision obtained with a convex lens used as a hand or stand magnifier is dependent upon the size or aperture of the lens, and on the eye-lens distance. The greater the eye-lens distance, the smaller the field of vision. The magnifying power (MP) of an optical system can be defined as It can be shown mathematically that where F e is the power of the eye-piece lens in dioptres, and F o is the power of the objective lens in dioptres. The practical usefulness of optical magnifying devices as low vision aids is limited by the following factors. 1. High magnification results in a reduced field of view, which makes rapid scanning of a line or page of print impossible. This factor also limits the usefulness of a distance low vision aid. 2. The object to be viewed has to be held close to the eye. 3. Magnification means that depth of focus is reduced. Thus the object lens distance is critical. 97

98 Spectacle Magnification Lens shape affects retinal image size. For our discussion, shape is defined as front surface power, thickness, and index of refraction. All these contribute to the shape factor magnification produced by a lens. The change in retinal image size due to lens refractive power is referred to as the power factor. Spectacle magnification (M spect ) can be expressed as follows: M spect = (M power ) (M shape ) where M power is the power factor, and Mshape is the shape factor. The power factor is calculated with the following formula: where d is the vertex distance and F v is the back vertex power. The formula for the shape factor is where t is the lens thickness, n is the lens s index of refraction, and F 1 is the power of the front surface. 98

99 Spectacle Magnification 99 Putting this all together, we have or M spect = (M power )(M shape ) Example: A polycarbonate lens has a power of D and a front surface refractive power of D. If the lens has a center thickness of 4.0 mm and the vertex distance is 14 mm, what is the magnification produced by the lens? It s straightforward to substitute in the formula for spectacle magnification as follows: M spect = (1.08)(1.01) = 1.09 x The spectacle magnification produced by the lens is Of this, 1.08 is due to the power of the lens and 1.01 is due to its shape.

100 Spectacle Magnification 100 Example: If a lens has a power of DS, but has a front surface power of DS. We ll assume that the lens is made of the same material and has the same center thickness and vertex distance. What magnification is produced by this lens? M spect = (1.08)(1.04) = 1.12 When two lenses of equal power are made of the same material and have the same thickness and vertex distance, the lens with the more curved front surface will produce greater magnification.

101 Spectacle Prescription The doctor specifies the optical characteristics a pair of ophthalmic lenses should provide for a given wearer by writing a spectacle prescription, or Rx. The Rx describes the focal powers needed to correct the refractive errors for distance and/or near vision, as well as any prescribed prism. Spectacle prescriptions are specified by eye OD (Oculus Dexter) or RE represents the wearer s right eye. OS (Oculus Sinister) or LE represents the wearer s left eye. OU (Oculi Uterque) when used represents a prescription suitable for both eyes. Table shows a common format for eyeglass prescriptions, including the sphere power (sph), cylinder power (cyl), axis of the cylinder, prism magnitude and orientation, and/or add power for the right (OD) and left (OS) eyes. TABLE: A typical prescription The spectacle Rx above depicts a right (OD) eye that requires a D lens for a myopia correction. The left (OS) eye requires a DS DC lens 135 for a compound myopic astigmatism correction. Both eyes (OU) require a D add power. No prism was prescribed. 101

102 The sphere (sph) power, cylinder (cyl) power, and axis describe the necessary focal powers required by the wearer for clear distance vision. 102

103 Questions A relaxed crystalline lens has a refractive index n = 1.42 and radii of curvature R 1 = mm, R 2 = mm. The lens is surrounded by two media of index Calculate the focal length and optical power of the lens, treating it as a thin lens. 2. An object is placed 10 cm in front of the cornea. (a) What is the image distance for the image formed by the cornea alone? (b) The image formed by the cornea serves as an object for the lens. Treat the lens as a thin lens 5 mm behind the cornea. Find the optical power of the lens necessary to form an image on the retina, 19 mm from the center of the lens. 3. Find the power of the lens necessary to correct an eye with a far point of (a) 25 cm; (b) 50 cm. 4. Find the far point of an eye for which a prescribed lens has an optical power of (a) d; (b) d. 5. What is the minimum power lens prescription that would allow an eye with a near point of 60 cm to see a clear image of a newspaper held 25 cm from the eye? 6. A contact lens has an optical power of d. What is the uncorrected far point of a person for whom a pair of such lenses is prescribed? 7. A certain eye, when relaxed, needs a 0.50 d lens in front of the eye to form a clear image on the retina of a distant object. Where is the (virtual) object seen by the eye?

104 Questions 1. A nearsighted person, wearing identical contact lenses, has a corrected range of clear vision from 20 cm to. The person s uncorrected near point is 10 cm. (a) What is the power of each lens? (b) What is the person s uncorrected range of clear vision? 2. A nearsighted man has a near point of 10 cm and a far point of 50 cm. He shaves without using his glasses. In order to see a clear image, what are the minimum and maximum distances of the mirror from his face? 3. Find the height of the retinal image of a person 1.5 m all, standing 4.0 m away. 4. (a) What is the angle subtended by an object whose image just covers the fovea, 0.4 mm in diameter? (b) At what distance would the image of a person s face, of height 20 cm, just fill the fovea? 5. Two laser beams fall on a screen and illuminate small circular areas. What is the minimum distance between the centers of the circles in order for an observer with normal visual acuity to be able to distinguish them from a distance of 10 m? 6. (a) What power lens would you prescribe for a patient with a far point of 35.0 cm? Neglect the eye lens distance. (b) Repeat, assuming an eye-corrective lens distance of 2.00 cm. Answer (a) 2.86 diopters (b) 3.03 diopters. 7. Suppose a lens is placed in a device that determines its power as 2.75 diopters. Find (a) the focal length of the lens and (b) the minimum distance at which a patient will be able to focus on an object if the patient s near point is 60.0 cm. Neglect the eye lens distance. Answers (a) 36.4 cm (b) 22.7 cm 104

105 TEŞEKKÜRLER Prof.Dr.A.Necmeddin YAZICI University Of Gaziantep, Optic and Acoustic Engineering 105